1

Came across the following exercise in Bartle's Elements of Real Analysis.

Let $X= (x_n)$ be a sequence of strictly positive real numbers, let $\lim \left({ \dfrac{x_{n + 1}}{x_n}}\right) = L$, and let $0 \lt \epsilon \lt L$.

Show that there exists $A \gt 0$ , $B \gt 0$ and $K \in \Bbb N$ such that $A(L - \epsilon)^n \le x_n \le B(L + \epsilon)^n$ for $n \ge K$. Then show that $\lim (\sqrt[n]{x_n}) = L$.

I think I've arrived at a solution but I'm sure there is a better way and this is what I seek. I would be grateful if someone could point me in the right direction.

The existence of $A = \dfrac {x_K}{(L - \epsilon)^K } \gt 0$ and $B = \dfrac {x_K}{(L + \epsilon)^K } \gt 0$ was easy enough to prove and now I need to prove the limit of $(\sqrt[n]{x_n})$ and I think I've taken a rather ugly route***.

For $n \ge K$

$$A(L - \epsilon)^n \le x_n \le B(L + \epsilon)^n$$

$$\sqrt[n]A(L - \epsilon) \le \sqrt[n]{x_n} \le \sqrt[n]B(L + \epsilon)$$

I had to do quite a few other things to prove that for positive $a$, $ \; \; \lim (a^{\frac 1 n}) = 1$ and hence by the above inequality arrive at $$\lim \left({\sqrt[n]A(L - \epsilon)}\right) \le \lim \left({\sqrt[n]{x_n}}\right) \le \lim \left({\sqrt[n]B(L + \epsilon)}\right)$$

$$ (L - \epsilon) \le \lim \left({\sqrt[n]{x_n}}\right) \le (L + \epsilon) $$ $$ \left|{\lim \left({\sqrt[n]{x_n}}\right) - L}\right| \le \epsilon $$

From here I argued that since the latter inequality can be made true for any $\epsilon \gt 0$ however small it must follow that $\lim \left({\sqrt[n]{x_n}}\right) = L$

Like I said I'm sure there is a better approach from the position marked ***. Would be grateful if someone could point me to it.

Martin Sleziak
  • 53,687
Ishfaaq
  • 10,034
  • 2
  • 30
  • 57
  • Some other posts on this site related to this result: http://math.stackexchange.com/questions/69386/inequality-involving-limsup-and-liminf, http://math.stackexchange.com/questions/76743/limit-of-fraca-n1a-n/ and http://math.stackexchange.com/questions/28476/finding-the-limit-of-n-sqrtnn/28487#28487 – Martin Sleziak Mar 21 '14 at 09:02
  • @Martin Sleziak: I'm sorry have not covered the Limit Superior yet so I'm guessing won't understand any of that. – Ishfaaq Mar 21 '14 at 09:05
  • If you have a look at the second question, the question is about limit (not limsup/liminf) and also one of the answers given there does not mention these two notions. In any case, I think that having links to other occurrences of questions about the same result is useful. (Even if it does not help you, it might help other people who come to your question.) – Martin Sleziak Mar 21 '14 at 09:08
  • @Martin Sleziak: I meant no criticism. Thanks for the links actually. Just letting you know that I don't think any of them contain the answer expected by this exercise. The answer you mentioned still uses another Theorem I have not heard of. The only one I think I might understand is an answer by t.b. in the third link. Having trouble tying this one with that though. – Ishfaaq Mar 21 '14 at 09:24
  • Well, what you explain IS the route to take, and it is not particularly "ugly". – Did Mar 23 '14 at 06:46
  • @Did: Really?? I'm a little unsure due to the fact that the inequalities hold only for $n \ge K$. Therefore having trouble wrapping up a concrete argument for the final limit. – Ishfaaq Mar 23 '14 at 15:11
  • Limits are only concerned with the terms such that $n\geqslant K$, whatever $K$ is, so what is the problem? (And yes, really.) – Did Mar 23 '14 at 16:23
  • @Did: Okay then. Guess I'll take your word. – Ishfaaq Mar 23 '14 at 16:52

1 Answers1

0

For $a>0$, $\lim (a^{1/n}) = \lim \exp \ln (a^{1/n}) = \lim \exp \frac{1}{n} \ln a \rightarrow \exp 0 = 1$.

Eric Towers
  • 67,037